Update, August 26, 2015: I’ve updated this article so that its code works with Swift 2.
This article expands and improves on it by showing you how to create i OS text fields that: In order to demonstrate this, I’ve created a quick sample app, Constrained Text Field Demo.
Before we get into the explanations, let me cut to the chase and just give you the code.
For the purposes of discussing constrained text fields, we need to consider only two files: I gave the outlets for the text fields sensible names, but I thought that it might be helpful to show you an annotated storyboard that points out which outlet belongs to which text field: The code in the view controller calls on some string utility methods that I decided to put into their own module: the String Utils.swift file: Let’s take a closer look at the code…
You can download it here [90K Xcode project and associated files, zipped]. It contains a set of text fields, each one with its own set of constraints: In this article, I’ll walk you through the app and show you how to create your own constrained text fields in i OS.
Better still, I’ll give you the project files so that you can experiment with the app.
between -128 and 255 inclusive is provided, it is interpreted as the ASCII value of a single character (negative values have 256 added in order to allow characters in the Extended ASCII range).
Any other integer is interpreted as a string containing the decimal digits of the integer.
Thank you (in anticipation) First, if you want a 6-digit password then the variable must be declared as 7 characters to allow room for the null-terminator that cin will add.
From my experience with teaching C (and several other languages), I predict that your biggest problem here will be with the use of pointers in the ¬_integer syntax.
Everyone, including myself, has to think quite a bit about how pointers work before they get it.
I still have the problem where it will not pick up if I have entered a letter.
for example, if I enter 1234k I will get an error message, I feel because it is not 6 characters long, but if I type 12345k It will allow it through because it is the right length even though it has a letter.